Posted by Tulela Haimbodi on Friday, August 10, 2012 at 8:31am.
mols CH3COOH = 0.1 x 0.045L = 0.00450
mols NaOH = 0.1 x 0.005 = 0.0005
.....NaOH + CH3COOH ==> CH3COONa + H2O
I..0.0005....0.0045.......0..........0
C.-0.0005...-0.0005....-0.0005...+0.0005
E....0.......0.0040.....0.0005....0.0005
(CH3COO^-) = mols/L = 0.0005/0.050 = ?
thanks DRBob....how can i find the amount of CH3COOH undissociated in the equilibrium mixture?
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